IMO Practice Test: Biomolecules | Class 12 Chemistry

Question 1 of 13

A sugar gives a positive Tollens' test but does not react with Schiff's reagent and shows mutarotation. The sugar is most likely:

sucrose

glucose

a non-sugar polysaccharide

an amino acid

Explanation: Glucose reduces Tollens' (open chain available) yet, being almost entirely cyclic, gives a negative Schiff's test and shows mutarotation.

Question 2 of 13

Equimolar hydrolysis of sucrose changes rotation from +66.5° to about −20° mainly because:

glucose is strongly laevorotatory

fructose is strongly laevorotatory

both products are dextrorotatory

water rotates light

Explanation: Fructose (−92°) outweighs glucose (+52.5°), making the invert-sugar mixture net laevorotatory.

Question 3 of 13

Two DNA samples X and Y melt (separate) at 75°C and 90°C respectively. Compared with X, sample Y is richer in:

A=T pairs

G≡C pairs

uracil

ribose

Explanation: G≡C pairs have three H-bonds versus two in A=T, so a higher G≡C content needs a higher melting temperature.

Question 4 of 13

At a pH equal to its isoelectric point, an amino acid placed in an electric field will:

move to the cathode

move to the anode

not migrate

decompose

Explanation: At the pI the net charge is zero (pure zwitterion), so there is no migration toward either electrode.

Question 5 of 13

Glycine has no D/L isomers because:

it is the smallest amino acid

its α-carbon is not a chiral centre

it is fat-soluble

it lacks a –COOH group

Explanation: In glycine the α-carbon carries two H atoms, so it has no four different groups and is not chiral — hence no optical isomers.

Question 6 of 13

A single DNA strand reads 5′-ATGC-3′. The complementary strand (5′→3′) is:

5′-TACG-3′

5′-GCAT-3′

5′-CGTA-3′

5′-ATGC-3′

Explanation: Pairing A–T, T–A, G–C, C–G gives the antiparallel complement, which read 5′→3′ is GCAT.

Question 7 of 13

Boiling does not change the molar mass measured for the peptide chain of a denatured protein because denaturation:

hydrolyses peptide bonds

leaves the primary structure intact

removes amino acids

adds water across each bond

Explanation: Denaturation breaks only the weak forces of the 2°/3° structure; the covalently bonded primary chain (and its mass) is unchanged.

Question 8 of 13

A patient stores a vitamin in the liver and develops toxicity on overdose. This vitamin is most likely:

vitamin C

vitamin B₁

vitamin A

vitamin B₁₂

Explanation: Fat-soluble vitamins (A, D, E, K) are stored in the liver/fat and can reach toxic levels; water-soluble ones are excreted.

Question 9 of 13

Cellulose and starch are both glucose polymers, yet only starch is digestible by humans. The key reason is the:

number of glucose units

presence of nitrogen

molar mass difference

type of glycosidic linkage (α vs β)

Explanation: Human amylase hydrolyses α-1,4 links (starch) but not β-1,4 links (cellulose).

Question 10 of 13

In protein synthesis the correct flow of genetic information is:

protein → mRNA → DNA

DNA → mRNA → protein

mRNA → DNA → protein

tRNA → DNA → mRNA

Explanation: DNA is transcribed to mRNA, which is translated on the ribosome to protein — the central dogma.

Question 11 of 13

Fructose forms a five-membered ring whereas glucose forms a six-membered ring because in fructose the ring closes between the:

C1 aldehyde and C5–OH

C2 keto group and C5–OH

C1 and C2

C6 and C1

Explanation: The C2 keto carbon of fructose reacts with the C5–OH to give a five-membered furanose ring; glucose's C1–CHO with C5–OH gives a six-membered pyranose ring.

Question 12 of 13

Which statement about Chargaff's rule for double-stranded DNA is correct?

A + G = C + T (purines = pyrimidines)

A = G and T = C

A + T = 0

all four bases are equal

Explanation: Because A pairs with T and G with C, the total purines (A+G) equal the total pyrimidines (C+T), and A=T, G=C individually.

Question 13 of 13

Maltose is reducing while sucrose is not. This is best explained by maltose having:

a free anomeric –OH (potential aldehyde)

an extra phosphate

a peptide bond

no glycosidic bond

Explanation: Maltose keeps one free anomeric carbon that can open to an aldehyde; sucrose locks both anomeric carbons, so it is non-reducing.

IMO Practice Test — Biomolecules